What is the general solution for cos(2x)-cos^2(x)=0 ?

The general solution for cos(2x)-cos^2(x)=0 is x=2pin,x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2x)-cos^2(x)=0

cos (2 x) - cos^{2} (x) = 0

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 x) - cos^{2} (x) = 0

Rewrite using trig identities

cos (2 x) - cos^{2} (x)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

cos^{2} (x) = cos (2 x) + sin^{2} (x)

= cos (2 x) - (cos (2 x) + sin^{2} (x))

Simplify

cos (2 x) - (cos (2 x) + sin^{2} (x)) : - sin^{2} (x)

cos (2 x) - (cos (2 x) + sin^{2} (x))

- (cos (2 x) + sin^{2} (x)) : - cos (2 x) - sin^{2} (x)

- (cos (2 x) + sin^{2} (x))

Distribute parentheses

= - (cos (2 x)) - (sin^{2} (x))

Apply minus-plus rules

+ (- a) = - a

= - cos (2 x) - sin^{2} (x)

= cos (2 x) - cos (2 x) - sin^{2} (x)

Add similar elements:

cos (2 x) - cos (2 x) = 0

= - sin^{2} (x)

= - sin^{2} (x)

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

\frac{- sin ^{2} ( x )}{- 1} = \frac{0}{- 1}

Simplify

sin^{2} (x) = 0

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

tan (π - θ) = tan (θ)

8 cos (x) = sec^{2} (x)

5 cot (θ) - 4 = 0

5 (sin (x) + 1) = 5

2 sin^{2} (t) - 3 cos (t) = 3

What is the general solution for cos(2x)-cos^2(x)=0 ?
The general solution for cos(2x)-cos^2(x)=0 is x=2pin,x=pi+2pin